STPS60170C High voltage power Schottky rectifier Datasheet - production data Description This dual diode Schottky rectifier is suited for high frequency switched mode power supplies. Packaged in TO-220AB this device is intended for use to enhance the reliability of the application. Table 1: Device summary Symbol Value IF(AV) 2 x 30 A VRRM 170 V T (max.) 175 C j VF (typ.) 0.76 V Features High junction temperature capability Good trade-off between leakage current and forward voltage drop Low leakage current Low thermal resistance Avalanche capability specified High frequency operation ECOPACK 2 compliant component January 2018 DocID11642 Rev 3 1/8 www.st.com This is information on a product in full production. Characteristics STPS60170C 1 Characteristics Table 2: Absolute ratings (limiting values, per diode, at 25 C, unless otherwise specified) Symbol Parameter Value Unit VRRM Repetitive peak reverse voltage 170 V I Forward rms current 45 A F(RMS) Per diode 30 Average forward current = 0.5, I T = 150 C A F(AV) C square wave Per device 60 tp = 10 ms I Surge non repetitive forward current 270 A FSM sinusoidal t = 10 s, p PARM Repetitive peak avalanche power 985 W T = 125 C j Tstg Storage temperature range -65 to +175 C (1) T Maximum operating junction temperature 175 j Notes: (1) (dP /dT ) < (1/R ) condition to avoid thermal runaway for a diode on its own heatsink. tot j th(j-a) Table 3: Thermal parameters Symbol Parameter Max. value Unit Per diode 1.0 Rth(j-c) Junction to case Total 0.7 C/W R Coupling 0.4 th(c) When the diodes 1 and 2 are used simultaneously: Tj (diode1) = P(diode1) x Rth(j-c) (per diode) + P(diode2) x Rth(c) Table 4: Static electrical characteristics (per diode) Symbol Parameter Test conditions Min. Typ. Max. Unit Tj = 25 C - 35 A (1) I Reverse leakage current V = V R R RRM T = 125 C - 8 35 mA j Tj = 25 C - 0.94 I = 30 A F T = 125 C - 0.72 0.76 j (2) VF Forward voltage drop V Tj = 25 C - 0.97 1.05 I = 60 A F T = 125 C - 0.86 0.92 j Notes: (1) Pulse test: tp = 5 ms, < 2% (2) Pulse test: tp = 380 s, < 2% To evaluate the conduction losses, use the following equation: 2 P = 0.60 x I + 0.0053 x I F(AV) F (RMS) 2/8 DocID11642 Rev 3